Q. A metal rod of resistance 20 Ω is fixed along a diameter of a conducting ring radius 0.1 m and lies on x-y plane. There is a magnetic field $ \displaystyle \vec{B} = 50 \hat{k} T $. The ring rotates with an angular velocity ω = 20 rod s-1 about its axis. An external resistance of 10 Ω is connected across the center of the ring and rim. The current through external resistance is
(a) 1/4
(b) 1/2
(c) 1/3
(d) 0
Ans:(c)
Sol: 
Potential difference between centre of the ring and the rim is
$ \displaystyle E = \frac{1}{2}B \omega r^2 $
$ \displaystyle E = \frac{1}{2}\times 50 \times 20 \times (0.1)^2 $
E = 5 volt
$ \displaystyle r = \frac{10 \times 10}{10 +10} $
r = 5 Ω
Current through external resistance
$ \displaystyle i = \frac{E}{R + r } = \frac{5}{10 + 5} $
i = 1/3 A